Question: You have found the following ages (in years) of 5 porcupines. Those porcupines were randomly selected from the 50 porcupines at your local zoo: $ 8,\enspace 2,\enspace 25,\enspace 14,\enspace 6$ Based on your sample, what is the average age of the porcupines? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 50 porcupines, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{8 + 2 + 25 + 14 + 6}{{5}} = {11\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {9} + {81} + {196} + {9} + {25}} {{5 - 1}} $ {s^2} = \dfrac{{320}}{{4}} = {80\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{80\text{ years}^2}} = {8.9\text{ years}} $ We can estimate that the average porcupine at the zoo is 11 years old. There is also a standard deviation of 8.9 years.